Zadatak 0990

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Ako je $$\cos 2\alpha =-\frac{63}{65}, \cos \beta =\frac{7}{\sqrt{130}}; \alpha ,\beta \in \left ( 0,\frac{\pi }{2} \right )$$ tada je \(\alpha +\beta \):

a) \(\frac{\pi }{4}\);          b) \(\frac{\pi }{3}\);         c) \(\frac{\pi }{2}\);         d) \(\frac{2\pi }{3}\);         e) \(\frac{3\pi }{4}\).

Rešenje:


Iz jednakosti $$\cos ^2\alpha =\frac{1+\cos 2\alpha }{2}=\frac{1}{65}, \alpha \in \left ( 0,\frac{\pi }{2} \right )$$

$$\Rightarrow \cos \alpha =\frac{1}{\sqrt{65}}$$

A onda je: $$\sin \alpha =+\sqrt{1-\cos ^2\alpha }=\sqrt{\frac{64}{65}}=\frac{8}{\sqrt{65}}$$

Slično, \(\beta \in \left ( 0,\frac{\pi }{2} \right )\) pa je:

$$\sin \beta =+\sqrt{1-\cos ^2\beta }=\sqrt{\frac{81}{130}}=\frac{9}{\sqrt{130}}$$

Iz adicionih formula dobijemo:

$$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $$

$$=\frac{1}{\sqrt{65}}\cdot \frac{7}{\sqrt{130}}-\frac{8}{\sqrt{65}}\cdot \frac{9}{\sqrt{130}}=-\frac{65}{\sqrt{65}\cdot \sqrt{130}}=-\frac{\sqrt{2}}{2}$$

$$\sin (\alpha +\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta $$

$$=\frac{8}{\sqrt{65}}\cdot \frac{7}{\sqrt{130}}+\frac{1}{\sqrt{65}}\cdot \frac{9}{\sqrt{130}}=\frac{65}{\sqrt{65}\cdot \sqrt{130}}=\frac{\sqrt{2}}{2}$$

Odatle imamo da je \(\alpha +\beta =\frac{3\pi }{4}\)

Tačan odgovor je e).